1. A man wearing a hat of extende length 12 cm is running in rain falling vertically downwards with speed 10m/s.The maximum speed with which man can run,so that rain drops do not fall on his face (the length of his face below the extended part of the hat is 16 cm) will be?

2. A taxi leaves the station X for station Y every 10 min. Simultaneously ,a taxi also leaves the station Y for station X every 10 min. The taxis move at the same constant speed and go frm X to Y or vice versa in 2 hrs . How many taxis comin from the other will meet each taxi moving from X to Y/

(1)

let hz. vel. of man is: V m/sec

rain dropes fall at vt. velocity: 10m/sec

tan x = V/10 = 12/16

V _man = 7.5 m/sec

(2)

Let distance btn. X and Y be S mts   and let V be constant speed of car

so S=7200V     (fom S=Vt relation )

Since each car leave with 10 mins difference : gap btn. each car is s=V*600 = S*600/7200 = S/12

                                                        gap btn. each car is = S/12 m

            following fig. explains it:

 X -------S₁-------S₂-------S₃-------S₄-------S₅-------S₆-------S₇------S₈-------S₉-------S₁₀-------S₁₁-------Y

case 1: So, when cars start simultaneously from X and Y

then first cars from X and Y meet at  S₆ and will start to cross 6 taxis till reaching X(orY), so it'll meet a total of 7 taxis including one at X(or Y)

case 2:

after 10 mins from departure of 1st car from X(or Y) , a car starting from Y (orX) will meet a total of 8 taxis including one at X(or Y)

the cases goes on till 2 hrs , and then a car from X(or Y) will meet a total of 13 taxis including one at X and Y.

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Since question 2 isn't clear for me, I have given all possible cases.

If u think , the answer is n't what u r looking for, plzz post the question clearly again .

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Regards,
Naga Ramesh
IIT Kgp - 2005 batch

Sir,

Sorry abt the wrong ques . The last line of the ques is:-

 How many taxis coming from the other side will meet each taxi moving from Y to X?